Hello Gang,
First, if you would like to performa an overall test of whether the
IQ interactions are necessary, you may find it most useful to use
anova to compare a full and reduced model. Something like:
ModelFit.full <-lme(mct~ IQ*age+IQ*I(age^2)+IQ*I(age^3), MyData,
random=~1|ID)
ModelFit.reduced <-lme(mct~ IQ + age+I(age^2)+I(age^3), MyData,
random=~1|ID)
anova(ModelFit.full, ModelFit.reduced, test="F")
Second, you don't have the syntax right for estimable(). As
described and shown by example in the manual page. The correct
syntax is:
library(gmodels)
estimable(ModelFit, c('IQ:age'=1, 'IQ:I(age^2) '= 1, 'IQ:I(age^3)' =
1))
Note the pattern of quoted name, followed by '=', and then the value
1 (not zero). This will perform a single joint test whether these
three coefficients are zero.
-G
On Oct 30, 2007, at 5:26PM , Gang Chen wrote:
> Dieter,
>
> Thank you very much for the help!
>
> I tried both glht() in multcomp and estimable() in gmodels, but
> couldn't get them work as shown below. Basically I have trouble
> specifying those continuous variables. Any suggestions?
>
> Also it seems both glht() and estimable() would give multiple t
> tests. Is there a way to obtain sort of partial F test?
>
>
>> ModelFit<-lme(mct~ IQ*age+IQ*I(age^2)+IQ*I(age^3), MyData,
> random=~1|ID)
>> anova(ModelFit)
>
> mDF denDF F-value p-value
> (Intercept) 1 257 54393.04 <.0001
> IQ 1 215 3.02 0.0839
> age 1 257 46.06 <.0001
> I(age^2) 1 257 8.80 0.0033
> I(age^3) 1 257 21.30 <.0001
> IQ:age 1 257 1.18 0.2776
> IQ:I(age^2) 1 257 0.50 0.4798
> IQ:I(age^3) 1 257 0.23 0.6284
>
>> library(multcomp)
>> glht(ModelFit, linfct = c("IQ:age = 0", "IQ:I(age^2) = 0", "IQ:I
> (age^3) = 0"))
> Error in coefs(ex[[3]]) :
> cannot interpret expression 'I''age^2' as linear function
>
>> library(gmodels)
>> estimable(ModelFit, rbind('IQ:age'=0, 'IQ:I(age^2) = 0', 'IQ:I
> (age^3) = 0'))
> Error in FUN(newX[, i], ...) :
> `param' has no names and does not match number of coefficients of
> model. Unable to construct coefficient vector
>
> Thanks,
> Gang
>
>
> On Oct 30, 2007, at 9:08 AM, Dieter Menne wrote:
>
>
>> Gang Chen <gangchen <at> mail.nih.gov> writes:
>>
>>
>>>
>>> Suppose I have a mixed-effects model where yij is the jth sample for
>>> the ith subject:
>>>
>>> yij= beta0 + beta1(age) + beta2(age^2) + beta3(age^3) + beta4(IQ) +
>>> beta5(IQ^2) + beta6(age*IQ) + beta7(age^2*IQ) + beta8(age^3 *IQ)
>>> +random intercepti + eij
>>>
>>> In R how can I get an F test against the null hypothesis of
>>> beta6=beta7=beta8=0? In SAS I can run something like contrast
>>> age*IQ
>>> 1, age^2*IQ 1, age^3 *IQ 1, but is there anything similar in R?
>>>
>>
>> Check packages multcomp and gmodels for contrast tests that work
>> with lme.
>>
>> Dieter
>>
>
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